How, to solve:
$$ \begin{align} \sum\limits_{k=1}^{\infty} \frac{1}{1-x^k} x^k = 0 \quad x \in \mathbb{R} \end{align} $$
Is it even possible?
There is a root at $x = 0$, but the graph shows that there should be also a root at around $x = -0.5454$.
Edit 1 (wrong): As comments said, my approach to solve was wrong:
Lets define $a := \frac{1}{1-x^k}$ then $$ \begin{align} \sum\limits_{k=1}^{\infty} a x^k = -a + \sum\limits_{k=0}^{\infty} a x^k = -a + \frac{a}{1-x} = \frac{(x-1)a + a}{1-x} = \frac{ax}{1-x} = 0 \end{align} $$
Because $a$ is an inverse, there should be only $x = 0$ a root. But this is not true.
You have $\sum_{k=1}^{\infty}\frac{x^k}{1-x^k}=\frac{\Psi_x(1)+\ln(1-x)}{\ln(x)}$, where $\Psi_x$ is the x-digamma function:
$\Psi_x(z):=\frac{1}{\Gamma_x(z)}\frac{\partial\Gamma_x(z)}{\partial z}$,
where $\Gamma_x$ is the x-gamma function (often called q-gamma function).
Valid for $-1<x<1$.